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Let $(M,g)$ be a Riemannian manifold and $g$ the Riemannian metric in coordinates $g=g_{\alpha \beta}dx^{\alpha} \otimes dx^{\beta}$, where $x^{i}$ are local coordinates on $M$. Denote by $g^{\alpha \beta}$ the inverse components of the inverse metric $g^{-1}$. Let $\nabla$ be the Levi-Civita connection of the metric $g$. Consider, locally, the function $\det((g_{\alpha \beta})_{\alpha \beta})$. It is known that $\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = 0$ by using normal coordinates etc...

I would like to show this fact without using normal coordinates. Just by computation. Here is what I have so far:

$$\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = \left [ g^{\gamma \delta} \partial_{\delta} \det((g_{\alpha \beta})_{\alpha \beta}) \right ] \partial_{\gamma} = \left [ \det((g_{\alpha \beta})_{\alpha \beta}) g^{\gamma \delta} g^{\beta \alpha} \partial_{\delta} g_{\alpha \beta}\right ] \partial_{\gamma}.$$

Here: the first equality sign follows from the definition of the gradient of a function and the second equality sign is the derivative of the determinant.

Question: How do I continue from here without using normal coordinates? Or are there any mistakes? If yes, where and which?

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  • $\begingroup$ Is $\det g$ really a well-defined function on $M$? $\endgroup$ Apr 14, 2019 at 10:43
  • $\begingroup$ It is not defined on the whole of $M$, only in a chart. $\endgroup$
    – Phillip
    Apr 14, 2019 at 11:10
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    $\begingroup$ I meant to say the claim that the gradient of det g is zero is false. You are requiring the function to be constant, but you have no restriction on g whatsoever. $g= x dx^2$ defined on some open interval of $\mathbb R$ will be a counterexample. $\endgroup$ Apr 14, 2019 at 11:28
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    $\begingroup$ It seems like you are confusing covariant derivative with gradient. These are two different concepts, even though both are denoted by $\nabla$. $\endgroup$ Apr 14, 2019 at 12:51
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    $\begingroup$ @Amitai I thaught that building the gradient of a function also gives you a tensor, a (1,0)-tensor. $\endgroup$
    – Phillip
    Apr 14, 2019 at 13:14

2 Answers 2

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Calculate $\Gamma^\nu_{\mu\nu}=\frac{1}{2}g^{\nu\kappa}(\partial_\mu g_{\nu\kappa}+\partial_\nu g_{\mu\kappa}-\partial_\kappa g_{\mu\nu})=\frac{1}{2}g^{\nu\kappa}\partial_\mu g_{\nu\kappa}$. Using the well known formula $$ \det A^{-1}\frac{d}{dt}\det A=\text{Tr}\left(A^{-1}\frac{d A}{dt}\right), $$ we obtain $$ \Gamma_\mu\equiv\Gamma^\nu_{\mu\nu}=\frac{1}{2}\frac{1}{g}\partial_\mu g=\frac{1}{2}\partial_\mu\ln|g|=\partial_\mu\ln \sqrt{|g|}, $$ where $g$ denotes the determinant of the matrix with elements $g_{\mu\nu}$.

The expression $\rho=\sqrt{|g|}$ transforms under a change of chart as follows: We have $g_{\mu^\prime \nu^\prime}=\partial_{\mu^\prime}x^\mu\partial_{\nu^\prime}x^\nu g_{\mu\nu}=J^\mu_{\mu^\prime}J^\nu_{\nu^\prime}g_{\mu\nu}$, taking determinants gives $$ g^\prime=\det J^2g \\ \sqrt{|g^\prime|}=|\det J|\sqrt{|g|}. $$ This object is called a "scalar density of weight 1". It makes sense in a coordinate-free manner too as a section of the density bundle, but whatever. One can show that the components of the covariant derivative of such an object is $$\nabla_\mu\rho=\partial_\mu\rho-\Gamma_\mu\rho=\partial_\mu\rho-\partial_\mu\ln\sqrt{|g|}\rho.$$ Inserting $\rho=\sqrt{|g|}$ gives $$ \nabla_\mu\sqrt{|g|}=\partial_\mu\sqrt{|g|}-\frac{1}{\sqrt{|g|}}\partial_\mu\sqrt{|g|}\sqrt{|g|}=\partial_\mu\sqrt{|g|}-\partial_\mu\sqrt{|g|}=0. $$

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  • $\begingroup$ Why the downvote? $\endgroup$ Oct 23, 2020 at 9:58
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One can give a short proof by considering the reciprocal basis $\partial^k=g^{k\sigma}\partial_{\sigma}$ and its derivative $\nabla_{\partial_k}\partial^j=-\Gamma^{j}{}_{ks}\partial^s$, because if we represent the metric tensor as $$g=g^{st}\partial_s\otimes\partial_t$$ then \begin{eqnarray*} g&=&g^{st}\partial_s\otimes\partial_t\\ &=&\partial_s\otimes g^{st}\partial_t\\ &=&\partial_s\otimes\partial^s. \end{eqnarray*}

So \begin{eqnarray*} \nabla_{\partial_k}g&=&\nabla_{\partial_k}(\partial_s\otimes\partial^s)\\ &=&(\nabla_{\partial_k}\partial_s)\otimes\partial^s+\partial_s\otimes\nabla_{\partial_k}\partial^s\\ &=&\Gamma^r{}_{ks}\partial_r\otimes\partial^s-\partial_s\otimes \Gamma^s{}_{kr}\partial^r\\ &=&\Gamma^r{}_{ks}\partial_r\otimes\partial^s-\Gamma^s{}_{kr}\partial_s\otimes \partial^r\\ &=&\Gamma^r{}_{ks}\partial_r\otimes\partial^s-\Gamma^r{}_{ks}\partial_r\otimes \partial^s\\ &=&0. \end{eqnarray*}

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  • $\begingroup$ I don't understand the downvote. $\endgroup$ Oct 23, 2020 at 10:28
  • $\begingroup$ @GiuseppeNegro: neither I do $\endgroup$
    – janmarqz
    Oct 23, 2020 at 16:39

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